ABCD-trapez isoscel
AD=BC
Fie CN_|_AB; CN-h
DM_|_AB,DM-h
MN=DC=18
AM=NB=(AB-MN)/2= (24-18)/2= 6/2=3
AN=AM+MN=3+18=21 cm
Fie AC_|_BC, mas<C=90 ⇒ ΔACB-dreptunghic
CN²=AN*NB
CN²=21*3=63 ⇔CN=√63 ⇒CN=3√7
a)A=(B+b)*h/2= (18+24)*3√7/2= 42*3√7/2= 21*3√7= 63√7
b)AC=BD
AC;BD-diag.
ΔANC,mas<N=90
AC²=AN²+NC²= 21²+ (3√7)= 441+63=504 ⇒AC=6√14
ΔBNC,mas<N=90
BC²=CN²+NB²= (3√7)²+ 3²= 63+9= 72 ⇒BC=6√2
c) fie CP_|_AD ⇒d(C;AD)=CP
A ΔADC=A ABCD -A ΔACB
ΔACB,mas<C=90
A ΔACB=AC*BC/2= (6√14*6√2)/2= 3√14*6√2= 18√28= 36√7
A ΔADC=63√7-36√7= 27√7
A ΔADC=b*h/2= CP*AD/2= CP*6√2/2= CP*3√2
CP*3√2=27√7
CP=27√7/3√2=
CP=9√7/√2 rationalizam
CP=9√14/2
